1. Problem 12 and 13, page 274, Chapter 7. Complete both the problems 12 and 13
ID: 467481 • Letter: 1
Question
1. Problem 12 and 13, page 274, Chapter 7. Complete both the problems 12 and 13 only for network 1 ONLY (given below) using
Hand calculations/normal tables – In an Excel spreadsheet, calculate the mean and variances of all activities and then separately (by hand) follow the methodology of duration risk assessments (PERT) to answer Question 13.//
Simulate the project network in @RISK will require you create a .mpp file in MS project – use modal times (m column below) to do that.
12. Given the immediate predecessors and a, m, b for each activity in the tables below, compute: a. te and V for each activity b. ES, EF, LS, and LF for each activity c. Te and Vp for the project. Activity Predecessors 8 D, E F, E 13. Refer to the first network in the above problem a. What is P(TeExplanation / Answer
Problem 12
a. Calculation of Expected time and variance of activities
Activity
Preceding activity
Optimistic time
Most Likely time
Pessimistic tie
Expected time
Variance
o
m
p
te =(o+4m+p)/6
2 =[(p – o)/6]2
A
-
7
9
11
9.00
0.44
B
A
1
2
3
2.00
0.11
C
A
7
8
9
8.00
0.11
D
B
2
5
11
5.50
2.25
E
C
2
3
4
3.00
0.11
F
C
1
4
8
4.17
1.36
G
D, E
6
7
8
7.00
0.11
H
F, E
2
6
9
5.83
1.36
b. Calculation of Earliest and latest times of activities by Forward and backward pass.
Activity
On Critical Path
Average Time
Earliest Start
Earliest Finish
Latest Start
Latest Finish
Total Slack
Te
ES
EF
LS
LF
LF - LS
A
y
9
0
9
0
9
0.00
B
n
2
9
11
13
15
3.50
C
y
8
9
17
9
17
0.00
D
n
6
11
17
15
20
3.50
E
y
3
17
20
17
20
0.00
F
y
4
17
21
17
21
0.00
G
y
7
20
27
20
27
0.00
H
y
6
21
27
21
27
0.00
Critical activities are activities with slack = 0. Thus critical activities are A, C, E, F, G, H.
Multiple Critical Path Exist:
Critical Path 1 = A-C-E-G
Critical Path 2 = A-C-F-H
Expected Critical Path duration = 27 days
Expected Critical Path Variance = Sum of variance of critical path
Critical Path 1(A-C-E-G) = 0.78
Critical Path 2 (A-C-F-H) = 3.28
Consider higher Variance among critical paths.
Expected Critical Path Standard deviation = square root of variance = 1.81 days
Thus,
c.
Expected Completion time of project = Te = 27 days
Expected project duration variance = Vp = 3.28 days
Expected project duration Standard deviation = = 1.81 days
Problem: 13
Te = 27 and = 1.81 days
a.
Here target completion time, X = 23 days
Z value for target time by following formula:
z = (X - µ)/
z = (23 - 27)/1.81 = -2.21
P(z = -2.21) = 0.013554
Likelihood that project can be completed in 23 days is 1.35%.
b.
Here target completion time, X = 32
Z value for target time by following formula:
z = (X - Te)/
z = (32 - 27)/1.81 = 2.762
P(z = 2.762) = 0.9971
Likelihood that project can be completed in 32 weeks is 99.71%.
c)
Number of days that would result in 99% probability of completion = 2310 Days
For the probability of 0.95, the Z score is 1.65, number of days that would result in 95% probability,
X = z x + Te = 1.65 x 1.81 + 27
X = 30 days.
Activity
Preceding activity
Optimistic time
Most Likely time
Pessimistic tie
Expected time
Variance
o
m
p
te =(o+4m+p)/6
2 =[(p – o)/6]2
A
-
7
9
11
9.00
0.44
B
A
1
2
3
2.00
0.11
C
A
7
8
9
8.00
0.11
D
B
2
5
11
5.50
2.25
E
C
2
3
4
3.00
0.11
F
C
1
4
8
4.17
1.36
G
D, E
6
7
8
7.00
0.11
H
F, E
2
6
9
5.83
1.36
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